Question

Find the equation of the normal to the solid 2x^2 −y^2 +8z^2 = 11 at a point where it
intersects the line x−3 = z =(y+1)÷2.

Answer 1

Solution:

$2x^2 −y^2 +8z^2 = 11$

$x−3 = z =(y+1)÷2$

they intersect at points:

$(\frac{5-\sqrt{7}}{3}, \frac{-11-2\sqrt{7}}{3}, \frac{-4-\sqrt{7}}{3})$

and

$(\frac{5+\sqrt{7}}{3}, \frac{-11+2\sqrt{7}}{3}, \frac{-4+\sqrt{7}}{3})$

Normals:

$\frac{x-\frac{5-\sqrt{7}}{3}}{4\cdot\frac{5-\sqrt{7}}{3}}=\frac{y-\frac{-11-2\sqrt{7}}{3}}{-2\cdot\frac{-11-2\sqrt{7}}{3}}=\frac{z-\frac{-4-\sqrt{7}}{3}}{16\cdot\frac{-4-\sqrt{7}}{3}}$

$\frac{x-\frac{5+\sqrt{7}}{3}}{4\cdot\frac{5+\sqrt{7}}{3}}=\frac{y-\frac{-11+2\sqrt{7}}{3}}{-2\cdot\frac{-11+2\sqrt{7}}{3}}=\frac{z-\frac{-4+\sqrt{7}}{3}}{16\cdot\frac{-4+\sqrt{7}}{3}}$